Is the Square Root of i^4 Equal to 1 or -1?Date: 02/24/2004 at 08:51:26 From: Mary Subject: complex numbers If you take the square root of i to the fourth power, does that equal i to the second power, which is equivalent to -1? Or can you simplify under the radical first and say i to the fourth power is 1 and the square root is then 1? Which approach is correct? Date: 02/24/2004 at 10:37:01 From: Doctor Jacques Subject: Re: complex numbers Hi Mary, I think the following answer from the Dr. Math archive will tell you what you want to know: http://mathforum.org/library/drmath/view/64430.html I found it by searching on these keywords: square root complex You can find other similar answers by doing the same search at http://mathforum.org/library/drmath/mathgrepform.html I hope this helps. Let me know if you'd like to talk about this some more, or if you have any other questions. - Doctor Jacques, The Math Forum http://mathforum.org/dr.math/ Date: 02/24/2004 at 10:42:12 From: Doctor Rob Subject: Re: complex numbers Thanks for writing to Ask Dr. Math, Mary! When you are dealing with roots of complex numbers, you must always take into consideration the multiple values which may be obtained. The square roots of i^4 are 1 and -1. On the other hand, if you take the square roots of i, you get (1 + i)/sqrt(2) and -(1 + i)/sqrt(2), and the fourth power of both of these is -1. These answers are not the same, but they do have a certain consistency. From this you can see that the rules you learned for working with radicals of nonnegative real numbers, such as sqrt(a*b) = sqrt(a)*sqrt(b) (a > 0, b > 0), sqrt(a/b) = sqrt(a)/sqrt(b) (a > 0, b > 0), do not necessarily hold for complex numbers or even negative real ones. (Try various combinations of 1 and -1 for a and b to see what I mean). For positive real numbers a, the symbol sqrt(a) is defined to be the positive square root of a. We call this the "principal value" of the square root of a. This is done partly to make the above rule work. For negative real numbers and complex numbers, there is no way to pick one of the two square roots in a consistent way as the principal value to make the above rules work. The best you can say is that whichever choices you make for square roots of such numbers, sqrt(a*b) = +/- sqrt(a)*sqrt(b), sqrt(a/b) = +/- sqrt(a)/sqrt(b) with the double sign symbol +/-. Sorry, but negative real numbers and complex numbers are just not as simple as positive real numbers when interacting with square root symbols! For nth roots of nonzero numbers, there are always n, of which at least n - 2 are complex. When a is real and positive, there are two real roots, and you can pick the positive real nth root to be the principal value. When a is real and negative, and n is odd, there is just one real root which is negative, and you can pick that negative real nth root to be the principal value. In all other cases, none of the nth roots are real numbers, and there is no good way to pick a principal value of the nth root of a in order to make the analogue of the first equation above work. When there is no principal value which can be assigned, the best one can do for the analogues of the above rules is to write (a*b)^(1/n) = w*a^(1/n)*b^(1/n), (a/b)^(1/n) = w*a^(1/n)/b^(1/n)m where w may be any one of the n nth roots of 1 in the complex numbers. Feel free to reply if I can help further with this question. - Doctor Rob, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/